Minimum Number of Changes to Make Binary String Beautiful | 2 Ways | Leetcode 2914 |codestorywithMIK
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This is the 49th Video of our Playlist "Binary Tree : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve an easy string Problem : Minimum Number of Changes to Make Binary String Beautiful | 2 Approaches | Leetcode 2914 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Minimum Number of Changes to Make Binary String Beautiful | 2 Approaches | Leetcode 2914 | codestorywithMIK
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My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/strings/Minimum Number of Changes to Make Binary String Beautiful.cpp
Leetcode Link : https://leetcode.com/problems/minimum-number-of-changes-to-make-binary-string-beautiful
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Summary :
Approach 1: Iterative Counting and Tracking
Description: This approach involves iterating through the string and counting consecutive characters while checking for matches with the previous character sequence. If the count of consecutive characters is even, the sequence is reset. If the count is odd, the character must be changed, and the change counter is incremented.
Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(1) because no additional data structures are used.
Key Points:
Tracks sequences of consecutive characters.
Ensures that changes are only made when the count of consecutive characters is odd.
Approach 2: Iterative Check in Pairs
Description: This simpler approach iterates through the string by checking characters in pairs (i.e., indexes i and i+1). If the two characters in a pair differ, a change is required, and the changes counter is incremented. The loop increments by 2 to ensure only adjacent pairs are checked.
Time Complexity: O(n), as it scans through the string in a single pass.
Space Complexity: O(1) since no extra space is used.
Key Points:
Operates by checking each adjacent pair directly.
Simpler logic focused on pairwise differences.
✨ Timelines✨
00:00 - Introduction
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